GCSE Maths · Algebra

Velocity–time graphs GCSE Questions and Worked Answers

A velocity–time graph shows speed with a direction sign at each time. Its gradient gives acceleration; signed area gives displacement. Add absolute areas to find total distance.

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Start with the meaning

What you need to know about velocity–time graphs

A trolley moving 3 metres to the right each second has speed 3 m/s. If we choose right as the positive direction, its velocity is +3 m/s. Moving equally fast to the left gives −3 m/s. Velocity includes direction; speed is the non-negative size of that velocity.

See the idea first

Each point records the velocity at one time

Time t goes along the horizontal axis; velocity v goes up the vertical axis. For example, the point (4, 8) means that at 4 seconds the trolley moves at +8 m/s. The line's height is velocity, while its slope describes how quickly velocity changes.

The trolley gains velocity, keeps a constant velocity, then slows to rest. The flat section at 8 m/s is motion, not a stop.

From problem to method

Typical problems you need to be able to solve

These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.

Read motion from a graph

What the problem asks: Describe a horizontal section at v = 8 m/s from t = 4 to t = 8 seconds.

How to solve it: The trolley continues in the positive direction at constant speed for four seconds. Its velocity does not change, so acceleration is zero.

Find acceleration

What the problem asks: Velocity rises uniformly from 2 to 10 m/s over four seconds. Find acceleration.

How to solve it: Velocity changes by 8 m/s in 4 seconds: acceleration = 8 ÷ 4 = 2 m/s². The square on seconds means a velocity change per second.

Find displacement or distance

What the problem asks: Travel at +3 m/s for four seconds, then −2 m/s for three seconds. Find both.

How to solve it: Positive displacement is 12 m, then negative displacement is −6 m. Net displacement is 6 m; total distance is 12 + 6 = 18 m. Area calculations are Higher-tier work.

A reliable routine

Use gradient or area for the quantity requested

For constant acceleration, the line's gradient is change in velocity ÷ elapsed time. For displacement, begin with a constant-velocity rectangle: width is time, height is velocity, so area is vt. Splitting a changing-velocity graph into thin strips leads to the same signed-area relationship.

Read both axis scales and units; identify positive and negative velocity.
For a straight-section acceleration, use (final velocity − initial velocity) ÷ time interval.
For Higher-tier displacement, split the area into rectangles, triangles or trapezia and count below-axis areas as negative.
For total distance, add the sizes of all areas. A curve may require an estimate using strips or a tangent.

Check: Negative acceleration does not always mean slowing down. If velocity is already negative and becoming more negative, speed is increasing.

Fully worked

Velocity–time graphs GCSE worked examples

Each solution explains the clue, why the method fits and how to check the result.

Example 1

Read a journey

2 marks

Question

Describe the three stages of the graph, including when the trolley is at rest.

The trolley gains velocity, keeps a constant velocity, then slows to rest. The flat section at 8 m/s is motion, not a stop.

From 0 to 4 s it speeds up uniformly from rest to 8 m/s. From 4 to 8 s it moves at constant 8 m/s. From 8 to 12 s it slows uniformly to rest. It is at rest at t = 0 and t = 12 s, not during the horizontal stage.

Example 2

Acceleration

2 marks

Question

Use the graph to find acceleration from 0 to 4 seconds.

The trolley gains velocity, keeps a constant velocity, then slows to rest. The flat section at 8 m/s is motion, not a stop.

a=\frac{8-0}{4-0}

=2\text{ m/s}^2

Its velocity increases by 2 m/s each second.

Example 3

Area of a complete journey

Higher only4 marks

Question

Find the distance travelled over the 12-second graph.

The trolley gains velocity, keeps a constant velocity, then slows to rest. The flat section at 8 m/s is motion, not a stop.

All velocities are non-negative, so distance equals area. First triangle: $\frac12(4)(8)=16$ m. Rectangle: $4(8)=32$ m. Final triangle: $\frac12(4)(8)=16$ m.

d=16+32+16=64\text{ m}

Tip: the units are m, not square metres, because m/s × s = m.

Example 4

A trapezium

Higher only3 marks

Question

Velocity increases uniformly from 4 m/s to 10 m/s over 6 seconds. Find displacement.

For a straight sloping segment, average velocity is the mean of its endpoints.

s=\frac{4+10}{2}\times6

=7\times6

=42\text{ m}

This is the area of the trapezium under the line.

Example 5

Reverse direction

Higher only4 marks

Question

A body moves at +4 m/s for 3 s, then at −2 m/s for 5 s. Ignore the instantaneous change in this simplified model. Find displacement and distance.

First signed area: $4(3)=12$ m. Second: $(-2)(5)=-10$ m.

\text{Displacement}=12-10=2\text{ m}

\text{Distance}=12+10=22\text{ m}

The body ends 2 m in the positive direction from its start.

Example 6

Estimate below a curve

Higher only4 marks

Question

A smooth velocity curve has readings 0, 3 and 8 m/s at times 0, 2 and 4 s. Use two straight-line strips (the trapezium rule) to estimate displacement.

Join adjacent readings by straight segments for the estimate. The first strip is a triangle because its initial velocity is zero; the same average-endpoints formula still works.

s_1=\frac{0+3}{2}\times2=3\text{ m}

s_2=\frac{3+8}{2}\times2=11\text{ m}

s\approx3+11=14\text{ m}

This is not exact because the actual graph is curved.

10 original questions · total 22 marks

Velocity–time graphs GCSE exam-style questions

Try each question before opening its fully worked answer. The difficulty rises through the set.

Before you startAllow about 27 minutes · Show your working. Give exact answers unless a question asks you to round, and include units where needed. · answers start collapsed

Constant velocity

1 mark

What does a flat velocity–time line at −3 m/s mean?

Show worked answer

Constant speed 3 m/s in the chosen negative direction. Acceleration is zero.

Rest

1 mark

At what velocity is an object stationary?

Show worked answer

At v = 0 m/s. A point where a sloping line crosses zero is an instant of rest, not necessarily a pause.

Increasing velocity

2 marks

Velocity increases uniformly from 5 to 17 m/s in 6 s. Find acceleration.

Show worked answer

a=\frac{17-5}{6}=\frac{12}{6}=2\text{ m/s}^2

Decreasing velocity

2 marks

Velocity falls uniformly from 12 to 3 m/s in 3 s. Find acceleration.

Show worked answer

a=\frac{3-12}{3}=-3\text{ m/s}^2

The velocity is positive but decreasing, so this object slows down.

Rectangle

Higher only2 marks

An object has constant velocity 7 m/s for 8 s. Find displacement.

Show worked answer

Area = height × width:

s=7\times8=56\text{ m}

Triangle

Higher only2 marks

A car accelerates uniformly from rest to 20 m/s in 5 s. Find distance.

Show worked answer

The area is a triangle:

d=\frac12\times5\times20=50\text{ m}

All velocity is non-negative.

Trapezium

Higher only3 marks

Velocity increases uniformly from 6 to 14 m/s over 10 s. Find displacement.

Show worked answer

s=\frac{6+14}{2}\times10

=10\times10=100\text{ m}

Crossing zero

Higher only4 marks

Velocity decreases linearly from +6 m/s at t = 0 to −6 m/s at t = 4 s. Find displacement and distance.

Show worked answer

The line crosses zero halfway, at t = 2 s. Each triangle has area $\frac12(2)(6)=6$ m. Signed areas cancel, so displacement is 0 m. Their sizes add, so distance is 12 m.

A tangent gradient

Higher only3 marks

A tangent to a velocity curve at t = 5 s passes through (3 s, 4 m/s) and (7 s, 12 m/s). Estimate acceleration at t = 5 s.

Show worked answer

Use two well-separated points on the tangent, not a chord of the curve.

a\approx\frac{12-4}{7-3}=2\text{ m/s}^2

The tangent gradient estimates the instantaneous rate.

Negative does not always mean slower

2 marks

Velocity changes from −2 m/s to −8 m/s in 3 s. Is the body speeding up or slowing down?

Show worked answer

Speed grows from 2 to 8 m/s, so it is speeding up in the negative direction. Acceleration is (−8 − (−2)) ÷ 3 = −2 m/s².

Examiner-style feedback

Common velocity–time graphs mistakes

Calling every flat line rest

A flat velocity graph means constant velocity. Only a flat line at zero means rest throughout.

Using signed area for total distance

Below-axis motion contributes positive distance even though its displacement is negative.

Using the angle of the drawn line

Gradient comes from axis values and units; it is not the apparent angle on the screen.

30-second recap

Choose, carry out, check

Translate the wording before you reach for a rule.

Height gives velocity.
Gradient gives acceleration.
Signed area gives displacement.
Absolute areas give total distance.

Quick answers

Velocity–time graphs FAQ

How is a speed–time graph different?

Speed is non-negative. Its area gives distance; a velocity graph may include negative values and signed displacement.

Can I use calculus?

It is not required here. GCSE questions use graph gradients, standard areas, trapezia and tangent estimates.

Build connected skills

What to revise next

Prerequisite rate

Speed, distance and time

Understand distance per unit time.

Revise speed, distance and time

Gradient

Straight-line graphs

Calculate rise divided by run.

Revise straight-line graphs

Content standards

Curriculum and rights review

A14/R14 motion interpretation and straight gradients; A15/R15 area calculations, trapezia and curved tangent estimates labelled Higher. Questions, suggested marks, solutions and diagrams are original practice material, not official past-paper questions.

Official specification references

AQA GCSE Maths subject content